Necessary and Sufficient Conditions for the Fractional Calculus of Variations

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2 Preliminaries 2.1 Review on fractional calculus There are several definitions of fractional derivatives and fractional integrals, like Riemann–Liouville, Caputo, Riesz, RieszCaputo, Weyl, Grunwald–Letnikov, Hadamard, Chen, etc. We will present the definitions of the first two of them. Except otherwise stated, proofs of results may be found in [18]. Let f :[a,b] → R be a function, α a positive real number, n the integer satisfying n − 1 ≤ α

n n x d − 1 d − − Dαf(x)= In αf(x)= (x − t)n α 1f(t)dt, a x dxn a x Γ(n − α) dxn a

2 and

n n n b d − (−1) d − − Dαf(x) = (−1)n In αf(x)= (t − x)n α 1f(t)dt, x b dxn x b Γ(n − α) dxn x respectively; 3. the left and right Caputo fractional derivatives of order α are deﬁned by

n x − d 1 − − C Dαf(x)= In α f(x)= (x − t)n α 1f (n)(t)dt, a x a x dxn Γ(n − α) a and

n b − d 1 − − C Dαf(x) = (−1)n In α f(x)= (−1)n(t−x)n α 1f (n)(t)dt, x b x b dxn Γ(n − α) x respectively. There exists a relation between the Riemann–Liouville and the Caputo fractional derivatives:

− n 1 (k) f (a) − C Dαf(x)= Dαf(x) − (x − a)k α a x a x Γ(k − α + 1) k =0 and − n 1 (k) f (b) − C Dαf(x)= Dαf(x) − (b − x)k α. x b x b Γ(k − α + 1) k =0 Therefore,

′ (n−1) C α α if f(a)= f (a)= ... = f (a)=0, then a Dx f(x)= aDx f(x) and

′ (n−1) C α α if f(b)= f (b)= ... = f (b)=0, then x Db f(x)= xDb f(x).

These fractional operators are linear, i.e.,

P(f(x)+ νg(x)) = Pf(x)+ ν Pg(x),

α α C α C α α α where P is aDx , xDb , a Dx , x Db , aIx or xIb , and and ν are real numbers. If f ∈ Cn[a,b], then the left and right Caputo derivatives are continuous on [a,b]. The main advantage of Caputo’s approach is that the initial conditions for fractional differential equations with Caputo derivatives take on the same form as for integer-order differential equations. Some properties valid for integer differentiation and integer integration re- main valid for fractional differentiation and fractional integration; namely the Caputo fractional derivatives and the Riemann–Liouville fractional integrals are inverse operations:

3 1. If f ∈ L∞(a,b) or f ∈ C[a,b], and if α> 0, then

C α α C α α a Dx aIx f(x)= f(x) and x Db xIb f(x)= f(x).

2. If f ∈ Cn[a,b] and if α> 0, then

n−1 f (k)(a) IαC Dαf(x)= f(x) − (x − a)k a x a x k! k =0 and n− 1 (−1)kf (k)(b) IαC Dαf(x)= f(x) − (b − x)k. x b x b k! k =0 We also need for our purposes integration by parts formulas. For α> 0, we have (cf. [4])

b b C α α g(x) a Dx f(x)dx = f(x) xDb g(x)dx a a − n 1 b α+j−n n−1−j + xDb g(x) xDb f(x) a j=0 and b b C α α g(x) x Db f(x)dx = f(x) aDx g(x)dx a a n−1 n+j α+j−n n−1−j b + (−1) aDx g(x) aDx f(x) a , j=0 k −k k −k where aDxg(x)= aIx g(x) and xDb g(x)= xIb g(x) if k< 0. Therefore, if 0 <α< 1, we obtain

b b C α α 1−α b g(x) a Dx f(x)dx = f(x) xDb g(x)dx + xIb g(x) f(x) a (1) a a and b b C α α 1−α b g(x) x Db f(x)dx = f(x) aDx g(x)dx − aIx g(x) f(x) a . (2) a a Moreover, if f is a function such that f(a)= f(b) = 0, we have simpler formulas:

b b C α α g(x) a Dx f(x)dx = f(x) xDb g(x)dx (3) a a and b b C α α g(x) x Db f(x)dx = f(x) aDx g(x)dx. (4) a a

4 Remark 1. As α goes to 1, expressions (1) and (2) reduce to the classical integration by parts formulas:

b b ′ ′ b g(x) f (x)dx = − f(x) g (x)dx + [g(x) f(x)]a a a and b b ′ ′ b g(x) (−f (x))dx = f(x) g (x)dx − [g(x) f(x)]a , a a respectively. Remark 2. Observe that the left member of equations (3) and (4) contains a Caputo fractional derivative, while the other one contains a Riemann–Liouville fractional derivative. However, since there exists a relation between the two derivatives, we could present this formula with only one fractional derivative, although in this case the resulting equation will contain some extra terms.

2.2 Fractional Euler–Lagrange equations From now on we ﬁx 0 <α,β < 1. Also, to simplify, we denote

C α C β [y](x) := (x, y(x), a Dx y(x), x Db y(x)).

Let

α β R C α C β a Eb = y :[a,b] → | a Dx y and x Db y exist and are continuous on [a,b] . Definition 2.1. The space of variations C V ar(a,b) for the Caputo derivatives is defined by C α β V ar(a,b)= h ∈ a Eb | h(a)= h(b)=0 . We now present first order necessary conditions of optimality for functionals, α β defined on a Eb , of the type

b J(y)= L[y](x)dx. (5) a We assume that the map (x,y,u,v) → L(x,y,u,v) is a function of class C1. Denoting by ∂iL the partial derivative of L with respect to the ith variable, i = 1, 2, 3, 4, we also assume that ∂3L has continuous right Riemann–Liouville fractional derivative of order α and ∂4L has continuous left Riemann–Liouville fractional derivative of order β. Deﬁnition 2.2. We say that y is a local minimizer (respectively local max- imizer) of J if there exists a δ > 0 such that J(y) ≤ J(y1) (respectively J(y) ≥ J(y1)) for all y1 such that y − y1 <δ.

5 In [1] Agrawal considers the problem of ﬁnding extremals for functionals containing left and right Riemann–Liouville fractional derivatives of the form

b α β J(y)= L(x, y(x), aDx y(x), xDb y(x))dx a and he derived an Euler–Lagrange equation for an extremum y of J, subject to the boundary conditions y(a)= ya, y(b)= yb:

α β ∂2L + xDb ∂3L + aDx ∂4L =0, for all x ∈ [a,b].

In [9] a new type of functional is studied, in case where the lower bound of the integral do not coincide with the lower bound of the fractional derivative:

B ∗ α J (y)= L(x, y(x), aDx y(x))dx, A where [A, B] ⊂ [a,b]. We also mention [3], where an Euler–Lagrange equation and a transversality condition are given, for functionals with left Caputo derivatives and a boundary condition on the initial point x = a. Theorem 2.3 ( [3]). Let J be the functional deﬁned by

b C α J(y)= L(x, y(x), a Dx y(x))dx. a

Let y be a local minimizer of J satisfying the boundary condition y(a) = y0. Then, y satisﬁes the following conditions:

α ∂2L + xDb ∂3L = 0 (6) and 1−α xIb ∂3L x=b =0. (7) Note that the Euler–Lagrange equation (6) contains a Riemann–Liouville fractional derivative, although J has only Caputo’s derivative. Also, the transversality condition (7), in general, contains fractional derivative terms. Thus, in order to solve a fractional variational problem, it may be required fractional boundary conditions. This result is then proven for functionals with higher order fractional derivatives. In [5] functionals with the left and right Caputo fractional derivatives are considered. We include here a short proof. For more on the subject we refer the reader to [11, 16, 23].

Theorem 2.4 ( [5]). Let J be the functional as in (5) and y a local minimizer of J satisfying the boundary conditions y(a)= ya and y(b)= yb. Then, y satisﬁes the Euler–Lagrange equation

α β ∂2L + xDb ∂3L + aDx ∂4L =0. (8)

6 Proof. Let ǫ be a small real parameter and η ∈ CV ar(a,b). Consider a variation of y; say y + ǫη. Since the Caputo derivative operators are linear, we have

b C α C α C β C β J(y + ǫη)= L(x, y + ǫη, a Dx y + ǫa Dx η, x Db y + ǫx Db η)dx. a We can regard J as a function of one variable, Jˆ(ǫ) = J(y + ǫη). Since y is the local minimizer, Jˆ attains an extremum at ǫ = 0. Diﬀerentiating Jˆ(ǫ) at zero, it follows that

b C α C β ∂2L η + ∂3L a Dx η + ∂4L x Db η dx =0. a Integrating by parts, and since η(a)= η(b) = 0, one ﬁnds that

b α β ∂2L + xDb ∂3L + aDx ∂4L η dx =0 a for all η ∈ C V ar(a,b). By the arbitrariness of η and by the fundamental lemma of the calculus of variations (see, e.g., [32, p. 32]), it follows that

α β ∂2L + xDb ∂3L + aDx ∂4L =0.

C β When the term x Db y is not present in the function L, then equation (8) reduces to a simpler one: α ∂2L + xDb ∂3L =0. Moreover, if we allow α = 1, and since in that case the right Riemann–Liouville fractional derivative is equal to −d/dx, we obtain the classical Euler–Lagrange equation: d ∂ L − ∂ L =0. 2 dx 3 3 Main results

Our ﬁrst contribution is to generalize Theorem 2.4 (cf. Theorem 3.1 below). We consider a new type of functional, where the lower bound of the integral is greater than the lower bound of the Caputo’s derivative, and the upper bound of the integral is less than the upper bound of the Caputo’s derivative. Because of this, we can not apply directly the integration by parts formula and some technical auxiliary procedures are required. A similar problem is addressed for the Riemann–Liouville fractional derivatives in [9].

7 3.1 The Euler–Lagrange equation We consider the functional B ∗ C α C β J (y)= L(x, y(x), a Dx y(x), x Db y(x))dx, (9) A where (x,y,u,v) → L(x,y,u,v) ∈ C1 and [A, B] ⊂ [a,b]. Theorem 3.1. If y is a local minimizer of J ∗ given by (9), satisfying the boundary conditions y(a)= ya and y(b)= yb, then y satisﬁes the system α β ∂2L + xDB∂3L + ADx ∂4L =0 for all x ∈ [A, B] Dα ∂ L − Dα ∂ L =0 for all x ∈ [a, A]  x B 3 x A 3 .  β β  ADx ∂4L − BDx ∂4L =0 for all x ∈ [B,b] Proof. Let ybe a minimizer and lety ˆ = y+ǫη be a variation of y, η ∈ CV ar(a,b), such that η(A) = η(B) = 0. Deﬁne the new function Jˆ∗(ǫ) = J ∗(y + ǫη). By hypothesis, y is a local extremum of J ∗ and so Jˆ∗ has a local extremum at ǫ = 0. Therefore, the following holds: B C α C β 0 = ∂2L η + ∂3L a Dx η + ∂4L x Db η dx A B B A C α C α = ∂2L ηdx + ∂3L a Dx ηdx − ∂3L a Dx ηdx A a a b b C β C β + ∂4L x Db ηdx − ∂4L x Db ηdx . A B Integrating by parts the four last terms gives:

B B A α α 0 = ∂2L ηdx + η xDB∂3Ldx − η xDA∂3Ldx A a a b b β β + η ADx ∂4Ldx − η BDx ∂4Ldx A B B A B A α α α = ∂2L ηdx + η xDB∂3Ldx + η xDB∂3Ldx − η xDA∂3Ldx A a A a B b b β β β + η ADx ∂4Ldx + η ADx ∂4Ldx − η BDx ∂4Ldx A B B A B α α α β = [xDB∂3L − xDA∂3L] ηdx + ∂2L + xDB∂3L + ADx ∂4L ηdx a A b β β + ADx ∂4L − BDx ∂4L ηdx. B Since η is an arbitrary function, we can assume that η(x) = 0 for all x ∈ [A, b] and so by the fundamental lemma of the calculus of variations, α α xDB∂3L − xDA∂3L =0, for all x ∈ [a, A].

8 Similarly, one proves the other two conditions:

α β ∂2L + xDB∂3L + ADx ∂4L =0, for all x ∈ [A, B], and β β ADx ∂4L − BDx ∂4L =0, for all x ∈ [B,b].

3.2 The fractional isoperimetric problem The isoperimetric problem is one of the most ancient problems of the calculus of variations. For example, given a positive real number l, what is the shape of the closed curve C of length l which deﬁnes the maximal area? The most essential contribution towards its rigorous proof was given in 1841 and is due to Jacob Steiner (1796–1863). We state the fractional isoperimetric problem as follows. Given a functional J as in (5), which functions y minimize (or maximize) J, when subject to given boundary conditions

y(a)= ya, y(b)= yb, (10) and an integral constraint

b I(y)= g[y]dx = l. (11) a 1 Here, similarly as before, we consider a function g of class C , such that ∂3g has continuous right Riemann–Liouville fractional derivative of order α and ∂4g has continuous left Riemann–Liouville fractional derivative of order β, and functions α β α β y ∈ a Eb . A function y ∈ a Eb that satisfies (10) and (11) is called admissible. Definition 3.2. An admissible function y is an extremal for I in (11) if it satisfies the equation

α β ∂2g[y](x)+ xDb ∂3g[y](x)+ aDx ∂4g[y](x)=0, for all x ∈ [a,b].

Observe that Deﬁnition 3.2 makes sense by Theorem 2.4. To solve the isoperimetric problem, the idea is to consider a new extended function. The exact formula for such extended function depends on y being or not an extremal for the integral functional I(y) (cf. Theorems 3.3 and 3.4). Theorem 3.3. Let y be a local minimum for J given by (5), subject to the conditions (10) and (11). If y is not an extremal for the functional I, then there exists a constant λ such that y satisﬁes

α β ∂2F + xDb ∂3F + aDx ∂4F = 0 (12) for all x ∈ [a,b], where F = L + λg.

9 C Proof. Let η1, η2 ∈ V ar(a,b) be two functions, ǫ1 and ǫ2 two reals, and consider the new function of two parameters

yˆ = y + ǫ1η1 + ǫ2η2. (13)

The reason why we consider two parameters is because we can choose one of them as a function of the other in order toy ˆ satisfy the integral constraint. Let

b Iˆ(ǫ1,ǫ2)= g[y + ǫ1η1 + ǫ2η2](x)dx − l. a It follows by integration by parts that

ˆ b ∂I C α C β = ∂2g η2 + ∂3g a Dx η2 + ∂4g x Db η2 dx ∂ǫ2 a (0,0) b α β = ∂2g + xDb ∂3g + aDx ∂4g η2dx. a We have assumed that y is not an extremal for I, and therefore there exists a function η2 satisfying the condition

∂Iˆ =0. (14) ∂ǫ2 (0,0) Using (14) and the fact that Iˆ(0, 0) = 0, by the Implicit Function Theorem there 1 exists a C function ǫ2(), deﬁned in some neighborhood of zero, such that

Iˆ(ǫ1,ǫ2(ǫ1))=0.

Therefore, there exists a family of variations of type (13) which satisfy the integral constraint. We will now prove condition (12). Similarly as before, we deﬁne a new function of two variables Jˆ(ǫ1,ǫ2) = J(ˆy). Since (0, 0) is a local minimum of Jˆ, subject to the constraint Iˆ(0, 0) = 0, and ∇Iˆ(0, 0) = 0, by the Lagrange Multiplier Rule (see, e.g., [32, p. 77]), there exists a constant λ for which the following holds: ∇(Jˆ(0, 0)+ λIˆ(0, 0)) = 0. Simple calculations show that

ˆ b ∂J α β = ∂2L + xDb ∂3L + aDx ∂4L η1dx ∂ǫ1 a (0,0) and ˆ b ∂I α β = ∂2g + xDb ∂3g + aDx ∂4g η1dx. ∂ǫ1 a (0,0) 10 In conclusion, it follows that

b α β α β ∂2L + xDb ∂3L + aDx ∂4L + λ ∂2g + xDb ∂3g + aDx ∂4g η1dx =0. a By the arbitrariness of η1 and the fundamental lemma of calculus of variations, one must have

α β α β ∂2L + xDb ∂3L + aDx ∂4L + λ ∂2g + xDb ∂3g + aDx ∂4g =0. This is equivalent to

α β ∂2F + xDb ∂3F + aDx ∂4F =0.

α Example 1. Let y(x) = Eα(x ), x ∈ [0, 1], where Eα is the Mittag–Leﬄer function: ∞ xk E (x)= , x ∈ R,α> 0. α Γ(αk + 1) k =0 When α = 1, the Mittag–Leﬄer function is the exponencial function, E1(x) = ex. The left Caputo fractional derivative of y is y (cf. [18, p. 98]),

C α 0 Dx y = y. Consider the following fractional variational problem:

1 C α 2 J(y)= (0 Dx y) dx → extr,  0 1 1 (15)  I(y)= y C Dαy dx = (y)2 dx,  0 x  0 0 y(0) = 1 and y(1) = Eα(1).   The augmented function is

C α C β C α 2 C α F (x, y, 0 Dx y, x D1 y, λ) = (0 Dx y) + λy 0 Dx y and the fractional Euler–Lagrange equation is

α β ∂2F + xD1 ∂3F + xD1 ∂4F =0 i.e., α C α xD1 (2 0 Dx y + λy)=0. A solution of this problem is λ = −2 and y = y. Observe that, as α → 1, the variational problem (15) becomes

1 ′ y 2 dx → extr, 0

11 y α =0.1

10 α =0.3

5 α =0.6 α =0.8 α =1 0.0 0.250.5 0.75 1.0 x

Figure 1: Solutions of problem (15).

1 ′ 1 yy dx = (e2 − 1), 2 0 y(0) = 1 and y(1) = e, and the Euler–Lagrange equation is

d d ′ ∂ F − ∂ F =0 ⇔− (2y − 2y)=0, (16) 2 dx 3 dx where F = y′2 − 2yy′. Also, for α = 1, y(x)= ex, which is obviously a solution of the diﬀerential equation (16) (cf. Figure 1). We now study the case when y is an extremal of I (the so called abnormal case). Theorem 3.4. Let y be a local minimum of J (5), subject to the conditions (10) and (11). Then, there exist two constants λ0 and λ, with (λ0, λ) = (0, 0), such that α β ∂2K + xDb ∂3K + aDx ∂4K =0 where K = λ0L + λg. Proof. Following the proof of Theorem 3.3, (0, 0) is an extremal of Jˆ subject to the constraint Iˆ = 0. Then, by the abnormal Lagrange multiplier rule (see, e.g., [32, p. 82]), there exist two reals λ0 and λ, not both zero, such that

∇(λ0Jˆ(0, 0)+ λIˆ(0, 0)) = 0.

12 Therefore, ∂Jˆ ∂Iˆ λ0 + λ =0. ∂ǫ1 ∂ǫ1 (0,0) (0,0) The rest of proof is similar to the one of Theorem 3.3. 3.3 An extension We now present a solution for the isoperimetric problem for functionals of type (9). Similarly, one has an integral constraint, but this time of the form

B ∗ C α C β I (y)= g(x, y(x), a Dx y(x), x Db y(x))dx = l. (17) A Again, we need the concept of extremal for a functional of type (17). Deﬁnition 3.5. A function y is called extremal for I∗ given by (17) if

α β ∂2g[y](x)+ xDB∂3g[y](x)+ ADx ∂4g[y](x)=0, for all x ∈ [A, B]. Theorem 3.6. If y is a local minimum of J ∗ given by (9), when restricted to ∗ the conditions y(a)= ya, y(b)= yb and (17), and if y is not an extremal for I , then there exists a constant λ such that α β ∂2F + xDB∂3F + ADx ∂4F =0 for all x ∈ [A, B] Dα ∂ F − Dα ∂ F =0 for all x ∈ [a, A]  x B 3 x A 3 (18)  β β  ADx ∂4F − BDx ∂4F =0 for all x ∈ [B,b] where F = L+ λg. Remark 3. In case [A, B] = [a,b], Theorem 3.6 is reduced to Theorem 3.3.

Proof. We consider a variation of formy ˆ = y + ǫ1η1 + ǫ2η2, where C η1, η2 ∈ V ar(a,b) and η1(A)= η1(B)= η2(A)= η2(B) = 0. Deﬁne Iˆ∗ by the expression

B ∗ Iˆ (ǫ1,ǫ2)= g[ˆy]dx − l. A Then, Iˆ∗(0, 0)=0 and

ˆ∗ B ∂I C α C β = ∂2g η2 + ∂3g a Dx η2 + ∂4g x Db η2 dx ∂ǫ2 A (0,0) A B α α α β = [xDB∂3g − xDA∂3g] η2dx + ∂2g + xDB∂3g + ADx ∂4g η2dx a A b β β + ADx ∂4g − BDx ∂4g η2dx B 13 (the last expression follows by integration by parts and some technical calculations as presented in the proof of Theorem 3.1). Let η2 be a function such that ∂Iˆ∗ =0 ∂ǫ2 (0,0) (its existence is guarantied since y is not an extremal for I∗). Therefore, we can 2 consider a subset of the family of functions {y + ǫ1η1 + ǫ2η2 | (ǫ1,ǫ2) ∈ R } that ∗ ∗ is admissible for the isoperimetric problem. Let Jˆ (ǫ1,ǫ2)= J (ˆy). Then, there exists a real λ such that

∇(Jˆ∗(0, 0)+ λIˆ∗(0, 0)) = 0. (19)

Similarly, one has

ˆ∗ A B ∂J α α α β = [xDB∂3L − xDA∂3L] η1dx + ∂2L + xDB∂3L + ADx ∂4L η1dx ∂ǫ1 a A (0,0) b β β + ADx ∂4L − BDx ∂4L η1dx, B and ˆ∗ A B ∂I α α α β = [xDB∂3g − xDA∂3g] η1dx + ∂2g + xDB∂3g + ADx ∂4g η1dx ∂ǫ1 a A (0,0) b β β + ADx ∂4g − BDx ∂4g η1dx. B By (19), ∂Jˆ∗ ∂Iˆ∗ + λ =0 ∂ǫ1 ∂ǫ1 (0,0) (0,0) and (18) follows from the arbitrariness of η1. The following result generalizes Theorem 3.4 and is proved in a similar way. Theorem 3.7. If y is a local minimum of J ∗ given by (9), subject to the boundary conditions (10) and the integral constraint (17), then there exist two constants λ0 and λ, not both zero, such that

α β ∂2K + xDB∂3K + ADx ∂4K =0 for all x ∈ [A, B] Dα ∂ K − Dα ∂ K =0 for all x ∈ [a, A]  x B 3 x A 3  β β  ADx ∂4K − BDx ∂4K =0 for all x ∈ [B,b]  where K =λ0L + λg.

14 3.4 Sufficient conditions of optimality We are now interested in finding sufficient conditions for J to attain local ex- tremes. Typically, some conditions of convexity over the Lagrangian are needed.

4 Deﬁnition 3.8. We say that f(x,y,u,v) is convex in S ⊆ R if ∂2f, ∂3f and ∂4f exist and are continuous, and the condition

f(x, y + y1,u + u1, v + v1) − f(x,y,u,v)

≥ ∂2f(x,y,u,v)y1 + ∂3f(x,y,u,v)u1 + ∂4f(x,y,u,v)v1 holds for every (x,y,u,v), (x, y + y1,u + u1, v + v1) ∈ S. Theorem 3.9. Suppose that the function L(x,y,u,v) is convex in [a,b] × R3. Then each solution y0 of the fractional Euler–Lagrange equation (8) minimizes

b J(y)= L[y]dx, a when restricted to the boundary conditions y(a)= y0(a) and y(b)= y0(b).

α β Proof. Let η ∈ a Eb be a function such that η(a) = η(b) = 0. Then, using integration by parts, it follows that

b J(y0 + η) − J(y0) = (L[y0 + η] − L[y0]) dx a b C α C β ≥ ∂2L[y0]η + ∂3L[y0]a Dx η + ∂4L[y0]x Db η dx a b α β = ∂2L + xDb ∂3L + aDx ∂4L [y0] η dx =0 a since y0 is a solution of (8). We just proved that J(y0 + η) ≥ J(y0). This procedure can be easily adapted for the isoperimetric problem. Theorem 3.10. Suppose that, for some constant λ, the functions L(x,y,u,v) and λg(x,y,u,v) are convex in [a,b] × R3. Let F = L + λg. Then each solution y0 of the fractional equation

α β ∂2F + xDb ∂3F + aDx ∂4F =0 minimizes b J(y)= L[y]dx, a under the constraints y(a)= y0(a) and y(b)= y0(b) and

b I(y)= g[y]dx = l, l ∈ R. a

15 Proof. Let us prove that y0 minimizes

b F (y)= (L[y]+ λg[y]) dx. a α β First, it is easy to prove that L(x,y,u,v)+ λg(x,y,u,v) is convex. Let η ∈ a Eb be such that η(a) = η(b) = 0. Then, by Theorem 3.9, F (y0 + η) ≥ F (y0). In α β other words, if y ∈ a Eb is any function such that y(a)= y0(a) and y(b)= y0(b), then b b b b L[y] dx + λg[y] dx ≥ L[y0] dx + λg[y0] dx. a a a a If we restrict to the integral constraint, we obtain

b b L[y] dx + l ≥ L[y0] dx + l, a a and so b b L[y] dx ≥ L[y0] dx, a a proving the desired result. Example 2. Recall Example 1. Since L(x,y,u,v) = u2 and λg(x,y,u,v) = −2yu are both convex, we conclude that y is actually a minimum for the fractional variational problem (15).

Acknowledgements

Work partially supported by the Centre for Research on Optimization and Con- trol (CEOC) from the “Funda¸cão para a Ciência e a Tecnologia” (FCT), cofi- nanced by the European Community Fund FEDER/POCI 2010.

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